Showing that $\lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\sum_{p<Q}1\right|=0$

Question

As the title of the question susggests, I would like to show that

The "trivial bound is that"

\begin{align*} \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|&\leq \lim_{Q\to\infty}\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left(\sum_{\substack{p<Q\\p|n}}p+\pi(Q)\right)\\ &=2 \end{align*}

where the equality is obtained by noting that $\frac{1}{p}$ numbers are multiplies of $p$, so the expected value of $\sum_{\substack{p<q \\ p|n}}p$ is exactly $\sum_{p<Q}1=\pi(Q)$. Thus, we are looking only for a "$o(\cdot)$" improvement. The first thought of mine would be to note that since $\sum_{\substack{p<q \\p|n}}p$ is an additive function and so by the Turan-Kubilius inequality

$$\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|^2\leq 4N\sum_{p<Q}p$$

The issue is, this inequality is worse than the trivial one since apply Cauchy-Shwartz we get that this inequality yields

\begin{align*} \frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|&\leq \frac{1}{\pi(Q)}\sqrt{\frac{1}{N}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|^2}\\ &\leq \frac{2}{\pi(Q)}\sqrt{\sum_{p<Q}p}\\ \end{align*}

where by the PNT this last term is on the order of $\sqrt{\frac{Q^2}{\log(Q)}}=\frac{Q}{\sqrt{\log(Q)}}$. This inequality would indicate to us that the sum would $\mathit{diverge}$, namely

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|=\Omega(\sqrt{\log(Q)})$$

I have proved that the sum can't go to zero too fast, and specifically that

$$\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|=\Omega(\log(Q)^{-\epsilon})$$

for any $\epsilon>0$. It does, however, feel natural that this sum should be at least $o(1)$.

Answer 1

It seems that the conjecture is false, if I did not miss some asymptotc issue The essence of what follows is that I show that the interior limit exceeds any function of $Q$ tending to $0$.

For any $t\geq 1$, denote by $p_t(Q)$ the density of those $n$ divisible by at least one $p$ with $t\pi(Q)<p<Q$. We have \begin{align*} p_t(Q)&=1-\prod_{t\pi(Q)<p<Q}\left(1-\frac1p\right) \sim 1-\exp\left(\sum_{t\pi(Q)<p<Q}\frac1p\right) \sim 1-\frac{\log Q}{\log (t\pi(Q))}\\ &\sim 1-\frac{\log Q}{\log Q+\log t-\log\log Q} =\frac{\log\log Q-\log t}{\log Q+\log t-\log\log Q}\\ &\sim \frac{\log\log Q-\log t}{\log Q}, \end{align*} where the estimates work in the regime $t\leq o(Q/\pi(Q))=o(\log Q)$. (More precisely, in any such regime this equivalence is uniform over $t\leq o(\log Q)$, as $q\to\infty$.)

Assume now that your conjecture is true, i.e., that $$ f(Q):=\lim_{N\to\infty}\frac{1}{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right| \to 0, \qquad Q\to\infty. $$ Take $N$ to be a multiple of $Q!$. Notice that at least $p_t(Q)N$ of the summands are larger than $(t-1)\pi(Q)$. Put $q=\log Q$. Summing up over $t=2,3,\dots,qf(Q)=o(q)$, we obtain \begin{align*} \frac1{N\pi(Q)}\sum_{n<N}\left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right| &\buildrel(*)\over\ge \sum_{t=2}^{qf(Q)} p_t(Q) \sim \sum_{t=2}^{qf(Q)} \frac{\log q-\log t}{q} =\frac1q\log\frac{q^{qf(Q)}}{(qf(Q))!}\\ &\sim \frac1q\log\left(\frac{qe}{qf(Q)}\right)^{qf(Q)} \sim f(Q)(-\log f(Q)). \end{align*} Since $f(Q)=o(1)$, the limit of the above expression as $N\to\infty$ cannot equal $f(Q)$.

Remark. The inequality $(*)$ holds, because if a number $n$ is accounted for $x$ times in the right sum, then its large prime divisor $p$ is larger than $(x+1)\pi(Q)$, so that $$ \left|\sum_{\substack{p<Q\\p|n}}p-\pi(Q)\right|>x\pi(Q). $$